Answer:
Greatest common factor (GCF) of 36 and 210 is 6.
Step-by-step explanation:
Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
Answer:
p value = 0.039
t = - 2.169
Step-by-step explanation:
Applying the null and alternate hypothesis

using excel worksheet to calculate for ( t and p )
t = -2.169
p = 0.039
from the results obtained
The conclusion is affected by the significance level because : 0.1 < p > 0.01
so when the significance level is = 0.1 the Null hypothesis is rejected and we can say the mean time interval will change while
if the significance level = 0.01 the Null hypothesis is accepted and we can not say the mean time interval has changed because the p -value is greater than 0.01
attached is the excel solution
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